The APL Orchard

10:22 AM

Any good solution to leetcode.com/problems/two-sum?

4 hours later…

2:15 PM

Subtract and compare?

@LdBeth Problem is to avoid using the same element twice.

The numbers are guaranteed unique?

no.

But there's a guarantee that there's only one answer.

2:30 PM

(n e. targt-n)#n NB. n numbers targt target value

you're supposed to return the indices, so I. instead of #n but this fails on e.g. targt=:6 and n=:3 2 4

⊃⍸(∘.<⍨⍳≢nums)∧target=∘.+⍨nums ?

Yes, that can work.

I tried starting with target-nums but couldn't get anywhere...

So you have {⊃⍸(∘.<⍨⍳≢⍵)∧⍺=∘.+⍨⍵}

I had {⊃(⍸⍺=∘.+⍨⍵)~2⍴¨⍋⍵}

I think you can write ∘.<⍨⍳≢⍵ as ∘.=⍨⍋⍵

No, that doesn't work.

2:46 PM

⍋ instead of ⍳≢ to save a character?

Yes. But also makes the code very obscure and less efficient.

With under, we could write {⍸0⍢(1 1∘⍉)⍺=∘.+⍨⍵}

Under can take an array left operand?

Under doesn't exist :-)

imo, @ should have been structural under, an it certainly does take an array operand as constant function.

3:01 PM

Would things like 0@(2∘|)⍳5 still work?

No. Current @ has implicit / or ⊇ (Select), but you'd need to do that yourself with Under, so it'd be 0@(2∘|⍛/)⍳5

3:24 PM

That seems very powerful, but I also like the simplicity of things like '-'@(=∘' ')

Speaking of which, how would you do '-'@(=∘' ') without @? I guess {(⍵,'-')[(⍳≢⍵)⌈(1+≢⍵)×' '=⍵]} works...

If you were going in the other direction, {⍵/⍨¯1*'-'=⍵} works :)

@Adám I.(d e. n)*.(d e. n#~ -. ~: n)+.(d ~: n) were t is target n is numbers and d is t-n

I tried to avoid outer product because that caps the complexity to O(n*2)

4:20 PM

@rabbitgrowth {r←⍵ ⋄ ((r=' ')/r)←'-' ⋄ r}

2 hours later…